Question

Question 2

Suppose a researcher is interested in the effectiveness in a new childhood exercise program implemented in a SRS of schools across a particular county. In order to test the hypothesis that the new program decreases BMI (Kg/m²), the researcher takes a SRS of children from schools where the program is employed and a SRS from schools that do not employ the program and compares the results. Assume the following table represents the SRSs of students and their BMIs.

Student Intervention Group	BMI (kg/m2)	Student Control Group	BMI (kg/m2)
A	18.6	A	21.6
B	18.2	B	18.9
C	19.5	C	19.4
D	18.9	D	22.6
		E	24.1
		F	23.6

A) Assuming that all the necessary conditions are met (normality, independence, etc.) carry out the appropriate statistical test to determine if the new exercise program is effective. Use an alpha level of 0.05. Do not assume equal variances.

B) Construct a 95% confidence interval about your estimate for the average difference in BMI between the groups.

Choos the most appropriate option

		Option 1 A) Ho: µ1=µ2 Ha: µ1<µ2 T statistic=13.141 Pvalue=0.001 Accept the Ho based on sufficient evidence at the alpha level of 0.05. B) Ci (15.17, 10.63)
		Option 2 A) Ho: µ1=µ2 Ha: µ1<µ2 T statistic=-3.141 Pvalue=0.01 Reject the Ho based on sufficient evidence at the alpha level of 0.05. Instead conclude that there is a statistically significant difference in mean BMI between the groups, with the control group having a consistently higher BMI on the average. B) Ci (-2.37, -1.36)
		Option 3 A) Ho: µ1=µ2 Ha: µ1<µ2 T statistic=-3.141 Pvalue=0.01 Reject the Ho based on sufficient evidence at the alpha level of 0.05. Instead conclude that there is a statistically significant difference in mean BMI between the groups, with the control group having a consistently higher BMI on the average. B) Ci (-5.17, -0.63)

Answer 1

Ho: µ1=µ2

Ha: µ1<µ2

Part A

To solve in excel, go to Data, select Data Analysis. Choose t-test: Two-Sample Assuming Unequal Variances.

t-Test: Two-Sample Assuming Unequal Variances
	Variable 1	Variable 2
Mean	18.80	21.70
Variance	0.30	4.66
Observations	4.00	6.00
Hypothesized Mean Difference	0.00
df	6.00
t Stat	-3.14
P(T<=t) one-tail	0.01
t Critical one-tail	1.94
P(T<=t) two-tail	0.02
t Critical two-tail	2.45

Since p-value (0.01) is less than 0.05, we reject null hypothesis and conclude that µ1<µ2. So, the new program does decreases BMI.

Part B: Confidence Intervals

Lower limit = (x1 - x2) - t*[((s12(n1-1) + s22(n2-1))/(n1+n2-2))^0.5*(1/n1+1/n2)^0.5]

= (18.8-21.7) - (-3.14)*(((0.3*(4-1)+4.66*(6-1))/(4+6-2))^0.5*(1/4+1/6)^0.5)

= 0.63

Upper limit = (x1 - x2) + t*[((s12(n1-1) + s22(n2-1))/(n1+n2-2))^0.5*(1/n1+1/n2)^0.5]

= (18.8-21.7) + (-3.14)*(((0.3*(4-1)+4.66*(6-1))/(4+6-2))^0.5*(1/4+1/6)^0.5)

= -6.43

Confidence interval = (-6.43,0.3)